4Sum

题目：

Given an array S of n integers, are there elements a,
b, c, and d in S such that a + b +
c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie,
a ≤ b ≤ c ≤ d)
The solution set must not contain duplicate quadruplets.

For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

A solution set is:
(-1,  0, 0, 1)
(-2, -1, 1, 2)
(-2,  0, 0, 2)

思想：
借用3Sum的思想，先固定住前两个数，然后在设定两个指针，根据sum的大小两个指针移动的位置。

代码：

#include <iostream>
#include <vector>
#include <algorithm>
#include <math.h>

using namespace std;

class Solution {
public:
vector<vector<int> > fourSum(vector<int>& nums, int target) {
vector<vector<int> >ret;
if(nums.size()<4) return ret;
sort(nums.begin(),nums.end());
int length=nums.size();
int sum;
for(int i=0;i<length-3;i++) //确定第一个数 a
{
if(i>0 &&nums[i]==nums[i-1])
{
//i++; //注意此处不要加1，否则会导致错误，满足条件的等式无法输出
continue;
}
int target2=target-nums[i]; //将4Sum转换成3Sum，target2=target-a;
for(int j=i+1;j<length-2;++j)
{
if(j>i+1 && nums[j]==nums[j-1])
{
// j++; //同上
continue;
}
int k=j+1;
int n=length-1;
vector<int> tmp;
while(k<n)
{
sum=nums[j]+nums[k]+nums
;

if(sum==target2)
{
if(k>j+1 && nums[k]==nums[k-1])
{
k++;
continue;
}
tmp.clear();
tmp.push_back(nums[i]);
tmp.push_back(nums[j]);
tmp.push_back(nums[k]);
tmp.push_back(nums
);
ret.push_back(tmp);
k++;//继续查找
}
else if(sum>target2)
{
n--;
}
else
{
k++;
}
}
}
}

return ret;
}
};

int main()
{
Solution s;
int arr[]={0,0,4,-2,-3,-2,-2,-3};
vector<int> ivec(arr,arr+8);
vector<vector<int> > ret;

ret=s.fourSum(ivec,-1);
vector<int> tmp;
for(auto iter=ret.begin();iter!=ret.end();++iter)
{
tmp.clear();
tmp=(*iter);
cout<<"("<<tmp[0]<<","<<tmp[1]<<","<<tmp[2]<<","<<tmp[3]<<")"<<endl;
}
system("pause");
return 0;
}